Post on 09-Aug-2020
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Changelog
Changes made in this version not seen in first lecture:12 September 2017: slide 28, 33: quote solution that uses z correctly
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last time
Y86 — choices, encoding and decoding
shift operatorsshr assembly, >> in Cright shift = towards least significant bitright shift = dividing by power of two
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on the quizzes in general
yes, I know quizzes are hard
intention: quiz questions from slides/etc. + some serious thought
(and sometimes I miss the mark)
main purpose: review material other than before exams, warningsign for me
why graded? because otherwise…
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on the quiz (1)
RISC versus CISC: about simplifying hardware
variable-length encoding is less simple for HW
instructions chosen more based on what’s simple for HW(e.g. push/pop not simple for HW)
more registers — simpler than adding more instructionscompensates for seperate memory instructions
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on the quiz (2)
instruction set — what the instructions do
size of memory address — operands to rmmovq, etc. mean what?
floating point support — do such instructions exist?
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constructing instructions
typedef unsigned char byte;byte make_simple_opcode(byte icode) {
// function code is fixed as 0 for nowreturn opcode * 16;// 16 = 1 0000 in binary
}
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constructing instructions in hardware
icode
0 0 0 0
opcode
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reversed shift right?
((((((((((((hhhhhhhhhhhhshr $-4, %reg
(((((((((((((hhhhhhhhhhhhhopcode >> (-4)
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shift left
x86 instruction: shl — shift leftC: value << amount
shl $amount, %reg (or variable: shr %cl, %reg)%reg (initial value)
%reg (final value)
1 0 1 1 0
0
0
0
1
1
…
…
…
…
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
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shift left
x86 instruction: shl — shift leftC: value << amount
shl $amount, %reg (or variable: shr %cl, %reg)%reg (initial value)
%reg (final value)
1 0 1 1 0
0
0
0
1
1
…
…
…
…
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
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left shift in math1 << 0 == 1 0000 00011 << 1 == 2 0000 00101 << 2 == 4 0000 0100
10 << 0 == 10 0000 101010 << 1 == 20 0001 010010 << 2 == 40 0010 1000
x << y = x × 2y
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left shift in math1 << 0 == 1 0000 00011 << 1 == 2 0000 00101 << 2 == 4 0000 0100
10 << 0 == 10 0000 101010 << 1 == 20 0001 010010 << 2 == 40 0010 1000
x << y = x × 2y
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extracting icode from more
11111000 0 0 1 0 0 0 0 0icode ifunrBrA
// % -- remainderunsigned extract_opcode1(unsigned value) {
return (value / 16) % 16;}
unsigned extract_opcode2(unsigned value) {return (value % 256) / 16;
}
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extracting icode from more
11111000 0 0 1 0 0 0 0 0icode ifunrBrA
// % -- remainderunsigned extract_opcode1(unsigned value) {
return (value / 16) % 16;}
unsigned extract_opcode2(unsigned value) {return (value % 256) / 16;
} 11
manipulating bits?
easy to manipulate individual bits in HW
how do we expose that to software?
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interlude: a truth table
AND 0 10 0 01 0 1
AND with 1: keep a bit the same
AND with 0: clear a bit
method: construct “mask” of what to keep/remove
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interlude: a truth table
AND 0 10 0 01 0 1
AND with 1: keep a bit the same
AND with 0: clear a bit
method: construct “mask” of what to keep/remove
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interlude: a truth table
AND 0 10 0 01 0 1
AND with 1: keep a bit the same
AND with 0: clear a bit
method: construct “mask” of what to keep/remove
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interlude: a truth table
AND 0 10 0 01 0 1
AND with 1: keep a bit the same
AND with 0: clear a bit
method: construct “mask” of what to keep/remove
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bitwise AND — &
Treat value as array of bits
1 & 1 == 1
1 & 0 == 0
0 & 0 == 0
2 & 4 == 0
10 & 7 == 2
… 0 0 1 0& … 0 1 0 0
… 0 0 0 0
… 1 0 1 0& … 0 1 1 1
… 0 0 1 0
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bitwise AND — &
Treat value as array of bits
1 & 1 == 1
1 & 0 == 0
0 & 0 == 0
2 & 4 == 0
10 & 7 == 2
… 0 0 1 0& … 0 1 0 0
… 0 0 0 0
… 1 0 1 0& … 0 1 1 1
… 0 0 1 0
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bitwise AND — &
Treat value as array of bits
1 & 1 == 1
1 & 0 == 0
0 & 0 == 0
2 & 4 == 0
10 & 7 == 2
… 0 0 1 0& … 0 1 0 0
… 0 0 0 0
… 1 0 1 0& … 0 1 1 1
… 0 0 1 0
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bitwise AND — C/assembly
x86: and %reg, %reg
C: foo & bar
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bitwise hardware (10 & 7 == 2)
10 7
...
0101 1110
0100
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extract opcode from larger
unsigned extract_opcode1_bitwise(unsigned value) {return (value >> 4) & 0xF; // 0xF: 00001111// like (value / 16) % 16
}
unsigned extract_opcode2_bitwise(unsigned value) {return (value & 0xF0) >> 4; // 0xF0: 11110000// like (value % 256) / 16;
}
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extract opcode from larger
extract_opcode1_bitwise:movl %edi, %eaxshrl $4, %eaxandl $0xF, %eaxret
extract_opcode2_bitwise:movl %edi, %eaxandl $0xF0, %eaxshrl $4, %eaxret
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more truth tables
AND 0 10 0 01 0 1
OR 0 10 0 11 1 1
XOR 0 10 0 11 1 0
&
conditionally clear bitconditionally keep bit
|
conditionally set bit
^
conditionally flip bit
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bitwise OR — |
1 | 1 == 1
1 | 0 == 1
0 | 0 == 0
2 | 4 == 6
10 | 7 == 15
… 1 0 1 0| … 0 1 1 1
… 1 1 1 1
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bitwise xor — ̂
1 ^ 1 == 0
1 ^ 0 == 1
0 ^ 0 == 0
2 ^ 4 == 6
10 ^ 7 == 13
… 1 0 1 0^ … 0 1 1 1
… 1 1 0 1
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negation / not — ~
~ (‘complement’) is bitwise version of !:
!0 == 1
!notZero == 0
~0 == (int) 0xFFFFFFFF (aka −1)
~2 == (int) 0xFFFFFFFD (aka −3)
~((unsigned) 2) == 0xFFFFFFFD
~ 0 0 … 0 0 0 01 1 … 1 1 1 1
32 bits
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negation / not — ~
~ (‘complement’) is bitwise version of !:
!0 == 1
!notZero == 0
~0 == (int) 0xFFFFFFFF (aka −1)
~2 == (int) 0xFFFFFFFD (aka −3)
~((unsigned) 2) == 0xFFFFFFFD
~ 0 0 … 0 0 0 01 1 … 1 1 1 1
32 bits
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negation / not — ~
~ (‘complement’) is bitwise version of !:
!0 == 1
!notZero == 0
~0 == (int) 0xFFFFFFFF (aka −1)
~2 == (int) 0xFFFFFFFD (aka −3)
~((unsigned) 2) == 0xFFFFFFFD
~ 0 0 … 0 0 0 01 1 … 1 1 1 1
32 bits
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note: ternary operator
w = (x ? y : z)
if (x) { w = y; } else { w = z; }
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one-bit ternary
(x ? y : z)
constraint: everything is 0 or 1
now: reimplement in C without if/else/||/etc.(assembly: no jumps probably)
divide-and-conquer:(x ? y : 0)(x ? 0 : z)
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one-bit ternary
(x ? y : z)
constraint: everything is 0 or 1
now: reimplement in C without if/else/||/etc.(assembly: no jumps probably)
divide-and-conquer:(x ? y : 0)(x ? 0 : z)
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one-bit ternary parts (1)
constraint: everything is 0 or 1
(x ? y : 0)
y=0 y=1x=0 0 0x=1 0 1
→ (x & y)
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one-bit ternary parts (1)
constraint: everything is 0 or 1
(x ? y : 0)
y=0 y=1x=0 0 0x=1 0 1
→ (x & y)
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one-bit ternary parts (2)
(x ? y : 0) = (x & y)
(x ? 0 : z)
opposite x: ~x
((~x) & z)
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one-bit ternary parts (2)
(x ? y : 0) = (x & y)
(x ? 0 : z)
opposite x: ~x
((~x) & z)
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one-bit ternary
constraint: everything is 0 or 1 — but y, z is any integer
(x ? y : z)
(x ? y : 0) | (x ? 0 : z)
(x & y) | ((~x) & z)
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multibit ternary
constraint: x is 0 or 1
old solution ((x & y) | (~x) & 1) only gets least sig. bit
(x ? y : z)
(x ? y : 0) | (x ? 0 : z)
((−x) & y) | ((−(x ^ 1)) & z)
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multibit ternary
constraint: x is 0 or 1
old solution ((x & y) | (~x) & 1) only gets least sig. bit
(x ? y : z)
(x ? y : 0) | (x ? 0 : z)
((−x) & y) | ((−(x ^ 1)) & z)
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constructing masks
constraint: x is 0 or 1
(x ? y : 0)
if x = 1: want 1111111111…1 (keep y)
if x = 0: want 0000000000…0 (want 0)
one idea: x | (x << 1) | (x << 2) | ...
a trick: −x
((-x) & y)
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constructing masks
constraint: x is 0 or 1
(x ? y : 0)
if x = 1: want 1111111111…1 (keep y)
if x = 0: want 0000000000…0 (want 0)
one idea: x | (x << 1) | (x << 2) | ...
a trick: −x
((-x) & y)
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two’s complement refresher
1−231
1+230
1+229
… 1+22
1+21
1+20
−1 =
−10
1
231 − 1−231
−231 + 10111 1111… 1111
1000 0000… 0000
1111 1111… 1111
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two’s complement refresher
1−231
1+230
1+229
… 1+22
1+21
1+20
−1 =
−10
1
231 − 1−231
−231 + 1
0111 1111… 1111
1000 0000… 0000
1111 1111… 1111
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two’s complement refresher
1−231
1+230
1+229
… 1+22
1+21
1+20
−1 =
−10
1
231 − 1−231
−231 + 10111 1111… 1111
1000 0000… 0000
1111 1111… 1111
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constructing masks
constraint: x is 0 or 1
(x ? y : 0)
if x = 1: want 1111111111…1 (keep y)
if x = 0: want 0000000000…0 (want 0)
one idea: x | (x << 1) | (x << 2) | ...
a trick: −x
((-x) & y)31
constructing other masks
constraint: x is 0 or 1
(x ? 0 : z)
if x = ��SS1 0: want 1111111111…1
if x = ��AA0 1: want 0000000000…0
mask: ��HH-x
−(x^1)
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constructing other masks
constraint: x is 0 or 1
(x ? 0 : z)
if x = ��SS1 0: want 1111111111…1
if x = ��AA0 1: want 0000000000…0
mask: ��HH-x −(x^1)
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multibit ternary
constraint: x is 0 or 1
old solution ((x & y) | (~x) & 1) only gets least sig. bit
(x ? y : z)
(x ? y : 0) | (x ? 0 : z)
((−x) & y) | ((−(x ^ 1)) & z)
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fully multibit
((((((((((((((hhhhhhhhhhhhhhconstraint: x is 0 or 1
(x ? y : z)
easy C way: !x = 0 or 1, !!x = 0 or 1x86 assembly: testq %rax, %rax then sete/setne(copy from ZF)
(x ? y : 0) | (x ? 0 : z)
((−!!x) & y) | ((−!x) & z)
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fully multibit
((((((((((((((hhhhhhhhhhhhhhconstraint: x is 0 or 1
(x ? y : z)
easy C way: !x = 0 or 1, !!x = 0 or 1x86 assembly: testq %rax, %rax then sete/setne(copy from ZF)
(x ? y : 0) | (x ? 0 : z)
((−!!x) & y) | ((−!x) & z)
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fully multibit
((((((((((((((hhhhhhhhhhhhhhconstraint: x is 0 or 1
(x ? y : z)
easy C way: !x = 0 or 1, !!x = 0 or 1x86 assembly: testq %rax, %rax then sete/setne(copy from ZF)
(x ? y : 0) | (x ? 0 : z)
((−!!x) & y) | ((−!x) & z)
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simple operation performance
typical modern desktop processor:bitwise and/or/xor, shift, add, subtract, compare — ∼ 1 cycleinteger multiply — ∼ 1-3 cyclesinteger divide — ∼ 10-150 cycles
(smaller/simpler/lower-power processors are different)
add/subtract/compare are more complicated in hardware!
but much more important for typical applications
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simple operation performance
typical modern desktop processor:bitwise and/or/xor, shift, add, subtract, compare — ∼ 1 cycleinteger multiply — ∼ 1-3 cyclesinteger divide — ∼ 10-150 cycles
(smaller/simpler/lower-power processors are different)
add/subtract/compare are more complicated in hardware!
but much more important for typical applications
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problem: any-bit
is any bit of x set?
goal: turn 0 into 0, not zero into 1
easy C solution: !(!(x))another easy solution if you have − or + (lab exercise)
what if we don’t have ! or − or +
how do we solve is x is two bits? four bits?
((x & 1) | ((x >> 1) & 1) | ((x >> 2) & 1) | ((x >> 3) & 1))
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problem: any-bit
is any bit of x set?
goal: turn 0 into 0, not zero into 1
easy C solution: !(!(x))another easy solution if you have − or + (lab exercise)
what if we don’t have ! or − or +
how do we solve is x is two bits? four bits?
((x & 1) | ((x >> 1) & 1) | ((x >> 2) & 1) | ((x >> 3) & 1))
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problem: any-bit
is any bit of x set?
goal: turn 0 into 0, not zero into 1
easy C solution: !(!(x))another easy solution if you have − or + (lab exercise)
what if we don’t have ! or − or +
how do we solve is x is two bits? four bits?
((x & 1) | ((x >> 1) & 1) | ((x >> 2) & 1) | ((x >> 3) & 1))
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wasted work (1)
((x & 1) | ((x >> 1) & 1) | ((x >> 2) & 1) | ((x >> 3) & 1))
in general: (x & 1) | (y & 1) == (x | y) & 1
(x | (x >> 1) | (x >> 2) | (x >> 3)) & 1
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wasted work (1)
((x & 1) | ((x >> 1) & 1) | ((x >> 2) & 1) | ((x >> 3) & 1))
in general: (x & 1) | (y & 1) == (x | y) & 1
(x | (x >> 1) | (x >> 2) | (x >> 3)) & 1
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wasted work (2)
4-bit any set: (x | (x >> 1) | (x >> 2) | (x >> 3)) & 1
performing 3 bitwise ors
…each bitwise or does 4 OR operations
2/3 of bitwise ORs useless — don’t use upper bits
...
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wasted work (2)
4-bit any set: (x | (x >> 1) | (x >> 2) | (x >> 3)) & 1
performing 3 bitwise ors
…each bitwise or does 4 OR operations
2/3 of bitwise ORs useless — don’t use upper bits
...
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any-bit: divide and conquer
four-bit input x = x1x2x3x4
(x >> 1) | x = (x1|0)(x2|x1)(x3|x2)(x4|x3) = y1y2y3y4
y2 = any-of(x1x2) = x1|x2, y4 = any-of(x3x4) = x3|x4
unsigned int any_of_four(unsigned int x) {int part_bits = (x >> 1) | x;return ((part_bits >> 2) | part_bits) & 1;
}
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any-bit: divide and conquer
four-bit input x = x1x2x3x4
(x >> 1) | x = (x1|0)(x2|x1)(x3|x2)(x4|x3) = y1y2y3y4
y2 = any-of(x1x2) = x1|x2, y4 = any-of(x3x4) = x3|x4
unsigned int any_of_four(unsigned int x) {int part_bits = (x >> 1) | x;return ((part_bits >> 2) | part_bits) & 1;
}
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any-bit: divide and conquer
four-bit input x = x1x2x3x4
(x >> 1) | x = (x1|0)(x2|x1)(x3|x2)(x4|x3) = y1y2y3y4
y2 = any-of(x1x2) = x1|x2, y4 = any-of(x3x4) = x3|x4
unsigned int any_of_four(unsigned int x) {int part_bits = (x >> 1) | x;return ((part_bits >> 2) | part_bits) & 1;
}
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parallelism
bitwise operations — each bit is seperate
same idea can apply to more interesting operations
010 + 011 = 101; 001 + 010 = 011 →01000001 + 01100010 = 10100011
sometimes specific HW supporte.g. x86-64 has a “multiply four pairs of floats” instruction
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parallelism
bitwise operations — each bit is seperate
same idea can apply to more interesting operations
010 + 011 = 101; 001 + 010 = 011 →01000001 + 01100010 = 10100011
sometimes specific HW supporte.g. x86-64 has a “multiply four pairs of floats” instruction
40
parallelism
bitwise operations — each bit is seperate
same idea can apply to more interesting operations
010 + 011 = 101; 001 + 010 = 011 →01000001 + 01100010 = 10100011
sometimes specific HW supporte.g. x86-64 has a “multiply four pairs of floats” instruction
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any-bit-set: 32 bits
unsigned int any(unsigned int x) {x = (x >> 1) | x;x = (x >> 2) | x;x = (x >> 4) | x;x = (x >> 8) | x;x = (x >> 16) | x;return x & 1;
}
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bitwise strategies
use paper, find subproblems, etc.
mask and shift(x & 0xF0) >> 4
factor/distribute(x & 1) | (y & 1) == (x | y) & 1
divide and conquer
common subexpression eliminationreturn ((−!!x) & y) | ((−!x) & z)becomesd = !x; return ((−!d) & y) | ((−d) & z)
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exercise
Which of these will swap last and second-to-last bit of anunsigned int x? (abcdef becomes abcdfe)/* version A */
return ((x >> 1) & 1) | (x & (~1));
/* version B */return ((x >> 1) & 1) | ((x << 1) & (~2)) | (x & (~3));
/* version C */return (x & (~3)) | ((x & 1) << 1) | ((x >> 1) & 1);
/* version D */return (((x & 1) << 1) | ((x & 3) >> 1)) ^ x;
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version A
/* version A */return ((x >> 1) & 1) | (x & (~1));// ^^^^^^^^^^^^^^// abcdef --> 0abcde -> 00000e
// ^^^^^^^^^^// abcdef --> abcde0
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^// 00000e | abcde0 = abcdee
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version B
/* version B */return ((x >> 1) & 1) | ((x << 1) & (~2)) | (x & (~3));// ^^^^^^^^^^^^^^// abcdef --> 0abcde --> 00000e
// ^^^^^^^^^^^^^^^// abcdef --> bcdef0 --> bcde00
// ^^^^^^^^^// abcdef --> abcd00
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version C
/* version C */return (x & (~3)) | ((x & 1) << 1) | ((x >> 1) & 1);// ^^^^^^^^^^// abcdef --> bcde00
// ^^^^^^^^^^^^^^// abcdef --> 00000f --> 0000f0
// ^^^^^^^^^^^^^// abcdef --> 0abcde --> 00000e
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version D
/* version D */return (((x & 1) << 1) | ((x & 3) >> 1)) ^ x;// ^^^^^^^^^^^^^^^// abcdef --> 00000f --> 0000f0
// ^^^^^^^^^^^^^^// abcdef --> 0000ef --> 00000e
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^// 0000fe ^ abcdef --> abcd(f XOR e)(e XOR f)
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int lastBit = x & 1;int secondToLastBit = x & 2;int rest = x & ~3;int lastBitInPlace = lastBit << 1;int secondToLastBitInPlace = secondToLastBit >> 1;return rest | lastBitInPlace | secondToLastBitInPlace;
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backup slides
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right shift in math1 >> 0 == 1 0000 00011 >> 1 == 0 0000 00001 >> 2 == 0 0000 0000
10 >> 0 == 10 0000 101010 >> 1 == 5 0000 010110 >> 2 == 2 0000 0010
x >> y =⌊x × 2−y⌋
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non-power of two arithmetic
unsigned times130(unsigned x) {return x * 130;
}
unsigned times130(unsigned x) {return (x << 7) + (x << 1); // x * 128 + x * 2
}
times130:movl %edi, %eaxshll $7, %eaxleal (%rax, %rdi, 2), %eaxret
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non-power of two arithmetic
unsigned times130(unsigned x) {return x * 130;
}unsigned times130(unsigned x) {
return (x << 7) + (x << 1); // x * 128 + x * 2}
times130:movl %edi, %eaxshll $7, %eaxleal (%rax, %rdi, 2), %eaxret
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non-power of two arithmetic
unsigned times130(unsigned x) {return x * 130;
}unsigned times130(unsigned x) {
return (x << 7) + (x << 1); // x * 128 + x * 2}
times130:movl %edi, %eaxshll $7, %eaxleal (%rax, %rdi, 2), %eaxret
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